Problem: Square ABCD is partitioned into nine congruent squares, with the center square partitioned again into nine congruent squares. What fractional part of square ABCD, shown here, is gray? Express your answer as a common fraction. [asy] fill((0,0)--(9,0)--(9,9)--(0,9)--cycle,gray); fill((9,9)--(12,9)--(12,12)--(9,12)--cycle,gray); fill((15,9)--(18,9)--(18,12)--(15,12)--cycle, gray); fill((12,12)--(12,15)--(15,15)--(15,12)--cycle,gray); fill((9,15)--(12,15)--(12,18)--(9,18)--cycle,gray); fill((15,15)--(18,15)--(18,18)--(15,18)--cycle,gray); fill((18,18)--(18,27)--(27,27)--(27,18)--cycle,gray); draw((0,0)--(27,0)--(27,27)--(0,27)--cycle,black+linewidth(1)); draw((9,0)--(9,27),black+linewidth(1)); draw((18,0)--(18,27),black+linewidth(1)); draw((0,9)--(27,9),black+linewidth(1)); draw((0,18)--(27,18),black+linewidth(1)); draw((12,9)--(12,18),black+linewidth(1)); draw((15,9)--(15,18),black+linewidth(1)); draw((9,12)--(18,12),black+linewidth(1)); draw((9,15)--(18,15),black+linewidth(1)); label("A",(0,27),NW); label("B",(0,0),SW); label("C",(27,0),SE); label("D",(27,27),NE); [/asy]
Explanation: There are nine total squares in the first partition, two of which are completely shaded gray and one of which is partially shaded gray. The square that is partially shaded gray is also divided into nine squares, five of which are shaded gray. Therefore, the shaded part of the partially shaded square is $\frac{1}{9}\cdot\frac{5}{9}=\frac{1\cdot5}{9\cdot9}=\frac{5}{81}$ of square ABCD. The squares that are completely shaded gray are $\frac{2}{9}$ of square ABCD. Therefore, the fractional part of square ABCD that is gray is $\frac{2}{9}+\frac{5}{81}=\frac{2\cdot9}{9\cdot9}+\frac{5}{81}=\frac{18}{81}+\frac{5}{81}=\boxed{\frac{23}{81}}$.